如何将JSON对象解析为TypeScript对象

moessi774:

我目前正在尝试将收到的JSON对象转换为具有相同属性的TypeScript类,但无法使其正常工作。我究竟做错了什么?

员工阶层

export class Employee{
    firstname: string;
    lastname: string;
    birthdate: Date;
    maxWorkHours: number;
    department: string;
    permissions: string;
    typeOfEmployee: string;
    note: string;
    lastUpdate: Date;
}

员工字符串

{
    "department": "<anystring>",
    "typeOfEmployee": "<anystring>",
    "firstname": "<anystring>",
    "lastname": "<anystring>",
    "birthdate": "<anydate>",
    "maxWorkHours": <anynumber>,
    "username": "<anystring>",
    "permissions": "<anystring>",
    "lastUpdate": "<anydate>"
    //I will add note later
}

我的尝试

let e: Employee = new Employee();

Object.assign(e, {
    "department": "<anystring>",
    "typeOfEmployee": "<anystring>",
    "firstname": "<anystring>",
    "lastname": "<anystring>",
    "birthdate": "<anydate>",
    "maxWorkHours": 3,
    "username": "<anystring>",
    "permissions": "<anystring>",
    "lastUpdate": "<anydate>"
});

console.log(e);

链接到打字稿游乐场

Nitzan Tomer:

编译器允许您将返回的对象强制JSON.parse转换为类的原因是因为typescript基于结构子类型
您实际上并没有的实例Employee,而是拥有一个具有相同属性的对象(如在控制台中看到的)。

一个简单的例子:

class A {
    constructor(public str: string, public num: number) {}
}

function logA(a: A) {
    console.log(`A instance with str: "${ a.str }" and num: ${ a.num }`);
}

let a1 = { str: "string", num: 0, boo: true };
let a2 = new A("stirng", 0);
logA(a1); // no errors
logA(a2);

操场上的代码

没有错误,因为a1满足类型A是因为它具有其所有属性,并且logA只要函数A具有相同的属性,即使它接收到的不是该实例的实例,也可以在运行时不会出现错误的情况下调用函数

当您的类是简单的数据对象并且没有方法时,这很好用,但是一旦您引入了方法,事情就会崩溃:

class A {
    constructor(public str: string, public num: number) { }

    multiplyBy(x: number): number {
        return this.num * x;
    }
}

// this won't compile:
let a1 = { str: "string", num: 0, boo: true } as A; // Error: Type '{ str: string; num: number; boo: boolean; }' cannot be converted to type 'A'

// but this will:
let a2 = { str: "string", num: 0 } as A;

// and then you get a runtime error:
a2.multiplyBy(4); // Error: Uncaught TypeError: a2.multiplyBy is not a function

操场上的代码


编辑

这很好用:

const employeeString = '{"department":"<anystring>","typeOfEmployee":"<anystring>","firstname":"<anystring>","lastname":"<anystring>","birthdate":"<anydate>","maxWorkHours":0,"username":"<anystring>","permissions":"<anystring>","lastUpdate":"<anydate>"}';
let employee1 = JSON.parse(employeeString);
console.log(employee1);

操场上的代码

如果您尝试JSON.parse在不是字符串的对象上使用它:

let e = {
    "department": "<anystring>",
    "typeOfEmployee": "<anystring>",
    "firstname": "<anystring>",
    "lastname": "<anystring>",
    "birthdate": "<anydate>",
    "maxWorkHours": 3,
    "username": "<anystring>",
    "permissions": "<anystring>",
    "lastUpdate": "<anydate>"
}
let employee2 = JSON.parse(e);

然后,您将得到错误信息,因为它不是字符串,而是对象,如果您已经以这种形式使用它,则无需使用JSON.parse

但是,正如我所写的那样,如果您采用这种方式,那么您将没有该类的实例,而只有一个具有与类成员相同属性的对象。

如果您想要一个实例,那么:

let e = new Employee();
Object.assign(e, {
    "department": "<anystring>",
    "typeOfEmployee": "<anystring>",
    "firstname": "<anystring>",
    "lastname": "<anystring>",
    "birthdate": "<anydate>",
    "maxWorkHours": 3,
    "username": "<anystring>",
    "permissions": "<anystring>",
    "lastUpdate": "<anydate>"
});

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